Given $f(x)\in\Bbb F_5[x]$ with $\deg f(x)=11$ for example. Suppose that $f(x)$ is irreducible. We know that the splitting field of $f(x)$ over $\Bbb F_5$ must exist, and its degree is $\le11!$. Generally speaking, $\Bbb F_{5^3}$ may not be big enough to contain all the roots of $f(x)$, and so does, say $\Bbb F_{5^{11}}$. So my question is that, what is the "biggest" possible order of the finite field we need to contains all of its root? I can only bound it by $5^{11!}$, but I think it may be strictly smaller. Are there some theorems related to this?
Update: Let $K$ be the splitting field of $f(x)$ over $\Bbb F_5$. Then by a theorem of finite field, $K=\Bbb F_5(u)$. If we can choose such $u$ as a root of $f(x)$, then since the degree of the minimal polynomial of $u$ is less then $11$, $[K:\Bbb F_5]\leq 11$. Then it seems that $\Bbb F_{5^{11}}$ is quite enough?
One knows that every extension of finite fields $\Bbb F_q\subset K$ is Galois with Galois group cyclic generated by the Frobenius morphism $x\mapsto x^q$.